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smokinjoe531 smokinjoe531
wrote...
Posts: 114
Rep: 0 0
11 years ago
I dont get how would you find the solutions or points that the parabola crosses the x axis.
ex-
Graph the quadratic equation 2x2 ? 4x ? 6 = 0 on your own graph paper. Choose the solution or solutions below that correspond to your graph. You may need to select more than one.

 x = -1

 x = -3

 x = 3

 x = 1

 No Solution
I thought it was 1 and -3 but apperently i was wrong. Please help me.
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datty117datty117
wrote...
Posts: 94
Rep: 1 0
11 years ago
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wrote...
11 years ago
If you plug in x = 1, you should get this:

= 2(1)^2 - 4(1) - 6
= 2 - 4 - 6
= -2 - 6
= -8
That is not zero, so x = 1 is not a solution.

If you plug in x = -3, you should get this:

= 2(-3)^2 - 4(-3) - 6
= 2(9) +12 - 6
= 18 + 12 - 6
= 30 - 6
= 24
This is not zero, so x = -3 is not a solution.

Just keep trying with the other values.
wrote...
11 years ago
x ² - 2x - 3 = 0

( x - 3 ) ( x + 1 ) = 0

x = 3 , x = - 1
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